Question: Let $f(x)=x^5+2x^3-x^2$. $f'(x)=$
According to the sum rule, the derivative of $x^5+2x^3-x^2$ is the sum of the derivatives of $x^5$, $2x^3$, and $-x^2$. The derivatives of these terms can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ For example, this is the derivative of the first term: $\dfrac{d}{dx}(x^5)=5x^4$ Here is the complete differentiation process: $\begin{aligned} &\phantom{=}f'(x) \\\\ &=\dfrac{d}{dx}(x^5+2x^3-x^2) \\\\ &=\dfrac{d}{dx}(x^5)+2\dfrac{d}{dx}(x^3)-\dfrac{d}{dx}(x^2)&&\gray{\text{Basic differentiation rules}} \\\\ &=5x^4+2\cdot3x^2-2x&&\gray{\text{The power rule}} \\\\ &=5x^4+6x^2-2x \end{aligned}$ In conclusion, $f'(x)=5x^4+6x^2-2x$.